Stuck on a Locus question (1 Viewer)

[narges-ts]

A point P (x,y) moves so that its distance from the point (2,3) and the line y= -1 is equal.
(i) Describe the locus of the P.
(ii) Write the equation of the locus of P.

 

[dan964]

A point P (x,y) moves so that its distance from the point (2,3) and the line y= -1 is equal.
(i) Describe the locus of the P.
(ii) Write the equation of the locus of P.

Geometrically, construct a perpendicular between (2,3) and the line y=-1. Take the midpoint of this line and construct a perpendicular to the midpoint (it should be parallel to the line y=-1). (It is possible to construct a line between (2,3) that is parallel to y=-1, that is the logic).

Numerically, calculate perpendicular distance between, and half it, let us call this LD. The line which is on the same side of the line y=-1 as the point (2,3) take a line LD units away from the line y=-1.

So your locus should a vertical [horizontal] line, parallel to y=-1.

I will let you complete the rest of part (i) and all of (ii)
ignore that, you can prove that the above answer is wrong by picking any other point and constructing a triangle, the length (hypothenuse) is longer than the distance of the length between the lines.

 

[Sp3ctre]

Geometrically, construct a perpendicular between (2,3) and the line y=-1. Take the midpoint of this line and construct a perpendicular to the midpoint (it should be parallel to the line y=-1). (It is possible to construct a line between (2,3) that is parallel to y=-1, that is the logic).

Numerically, calculate perpendicular distance between, and half it, let us call this LD. The line which is on the same side of the line y=-1 as the point (2,3) take a line LD units away from the line y=-1.

So your locus should a vertical line, parallel to y=-1.

I will let you complete the rest of part (i) and all of (ii)

Is the locus not a parabola?

 

[seanieg89]

Geometrically, construct a perpendicular between (2,3) and the line y=-1. Take the midpoint of this line and construct a perpendicular to the midpoint (it should be parallel to the line y=-1). (It is possible to construct a line between (2,3) that is parallel to y=-1, that is the logic).

Numerically, calculate perpendicular distance between, and half it, let us call this LD. The line which is on the same side of the line y=-1 as the point (2,3) take a line LD units away from the line y=-1.

So your locus should a vertical line, parallel to y=-1.

I will let you complete the rest of part (i) and all of (ii)

This is not correct, it is a parabola.

Direct calculation:
(x-2)^2+(y-3)^2=(y+1)^2
=> 8y=x^2-4x+12.

You could also immediately conclude that it is a parabola from the geometric definition of a parabola, and read off the focal length a by looking at the distance between the given focus and directrix.

(Also note that lines parallel to y=-1 are not vertical, they are horizontal.)

 

[dan964]

This is not correct, it is a parabola.

Direct calculation:
(x-2)^2+(y-3)^2=(y+1)^2
=> 8y=x^2-4x+12.

You could also immediately conclude that it is a parabola from the geometric definition of a parabola, and read off the focal length a by looking at the distance between the given focus and directrix.

(Also note that lines parallel to y=-1 are not vertical, they are horizontal.)

yeah my bad. brain is kind of fried at the moment.